Question

Evaluate the following integral:​

Answer 1

$\large\underline{\sf{Solution-}}$

Given integral is

$\rm :\longmapsto\:\displaystyle\int\rm \frac{ {x}^{e - 1} + {e}^{x - 1} }{ {x}^{e} + {e}^{x} } \: dx$

To evaluate this integral, we use Method of Substitution.

So, Substitute

$\rm :\longmapsto\: {x}^{e} + {e}^{x} = y$

$\rm :\longmapsto\: \dfrac{d}{dx}({x}^{e} + {e}^{x}) = \dfrac{d}{dx}y$

$\rm :\longmapsto\: {ex}^{e - 1} + {e}^{x} = \dfrac{dy}{dx}$

$\rm :\longmapsto\:e( {x}^{e - 1} + {e}^{x - 1}) = \dfrac{dy}{dx}$

$\rm :\longmapsto\:({x}^{e - 1} + {e}^{x - 1})dx = \dfrac{dy}{e}$

So, on substituting all these values in above integral, we get

$\rm \: = \: \displaystyle\int\rm \frac{dy}{e \: y}$

$\rm \: = \:\dfrac{1}{e} \displaystyle\int\rm \frac{dy}{ y}$

$\rm \: = \:\dfrac{1}{e}log |y| + c$

$\rm \: = \:\dfrac{1}{e}log \bigg| {x}^{e} + {e}^{x} \bigg| + c$

Hence,

$\rm :\longmapsto\:\boxed{\tt{ \displaystyle\int\rm \frac{ {x}^{e - 1} + {e}^{x - 1} }{ {x}^{e} + {e}^{x} } \: dx = \frac{1}{e}log \bigg| {x}^{e} + {e}^{x}\bigg| + c}}$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

LEARN MORE :-

$\begin{gathered}\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \displaystyle \int \rm \:f(x) \: dx\\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf kx + c \\ \\ \sf sinx & \sf - \: cosx+ c \\ \\ \sf cosx & \sf \: sinx + c\\ \\ \sf {sec}^{2} x & \sf tanx + c\\ \\ \sf {cosec}^{2}x & \sf - cotx+ c \\ \\ \sf secx \: tanx & \sf secx + c\\ \\ \sf cosecx \: cotx& \sf - \: cosecx + c\\ \\ \sf tanx & \sf logsecx + c\\ \\ \sf \dfrac{1}{x} & \sf logx+ c\\ \\ \sf {e}^{x} & \sf {e}^{x} + c\end{array}} \\ \end{gathered}\end{gathered}$